NDA Mathematics Question (2026) — Solution

Step-by-step solution

Given a b = c and b c = a . Taking the dot product of the first equation with c : ( a b ) c = c c [ a b c ] = | c |^2 Taking the dot product of the second equation with a : ( b c ) a = a a [ b c a ] = | a |^2 Using the properties of the scalar triple product, [ a b c ] = [ b c a ] = [ c a b ]. Thus, | a |^2 = | c |^2 = [ a b c ]. Evaluating Statement I: The left-hand side is ( a b ) c + ( b c ) a = [ a b c ] + [ b c a ] = 2[ a b c ]. The right-hand side is ( c a ) b = [ c a b ] = [ a b c ]. For the statement to be true, 2[ a b c ] = [ a b c ] [ a b c ] = 0, which implies | c |^2 = 0. This is not generally true for non-zero vectors. Hence, Statement I is incorrect. Evaluating Statement II: Substituting the given cross products into the expression: \ ( a b ) ( b c )\ b = ( c a ) b = [ c a b ] = [ a b c ] = | c |^2 This expression equals 1 only if | c | = 1. Since there is no condition restricting the magnitude of c to 1, Statement II is also incorrect. Therefore, neither Statement I nor Statement II is correct. Answer: Neither I nor II